Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

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The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

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Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long double esp = 1e-11;
const long double PI = acos((long double)-1);
const long long int INF = 0x3f3f3f3fll;
const int MOD = 1000000007ll;
const int maxn = 100;
int h[110][110];
int ans1[110], ans2[110], n, m;
int check(int t) {
  int ans = 0;
  for (int x = 1; x <= m; x++)
    if (h[1][x] < t)
      return -1;
    else
      ans2[x] = h[1][x] - t;
  for (int x = 1; x <= n; x++) {
    ans1[x] = h[x][1] - ans2[1];
    if (ans1[x] < 0) return -1;
    for (int y = 1; y <= m; y++) {
      if (h[x][y] - ans2[y] != ans1[x]) return -1;
    }
  }
  for (int x = 1; x <= m; x++) ans += ans2[x];
  for (int x = 1; x <= n; x++) ans += ans1[x];
  return ans;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int x = 1; x <= n; x++) {
    for (int y = 1; y <= m; y++) scanf("%d", &h[x][y]);
  }
  int flag = -1, ans = INF, p = -1, tem;
  for (int x = 0; x <= h[1][1]; x++)
    if ((tem = check(x)) != -1) {
      flag = 1;
      if (tem < ans) {
        ans = tem;
        p = x;
      }
    }
  if (flag == -1) {
    puts("-1");
    return 0;
  }
  ans = check(p);
  printf("%d\n", ans);
  for (int x = 1; x <= n; x++)
    while (ans1[x] > 0) printf("row %d\n", x), ans1[x]--;
  for (int x = 1; x <= m; x++)
    while (ans2[x] > 0) printf("col %d\n", x), ans2[x]--;
  return 0;
}